Closed Subset of Real Numbers with Lower Bound contains Infimum

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Theorem

Consider the real number line as a metric space under the usual metric.

Let $A \subseteq \R$ such that $A$ is closed in $\R$ and $A \ne \O$.

Let $A$ be bounded below.


Then $A$ contains its infimum.


Proof

From Infimum of Bounded Below Set of Reals is in Closure:

$\inf A \in \map \cl A$

From Set is Closed iff Equals Topological Closure:

$A = \map \cl A$

Therefore $\inf A \in A$.

$\blacksquare$


Sources